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Topic: Fractious Fractions
Replies: 10   Last Post: Jun 6, 2013 7:14 AM

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David C. Ullrich

Posts: 3,141
Registered: 12/13/04
Re: Fractious Fractions
Posted: Jun 5, 2013 12:20 PM
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On Tue, 4 Jun 2013 00:11:59 -0700, William Elliot <marsh@panix.com>
wrote:

>On Mon, 3 Jun 2013, dullrich@sprynet.com wrote:
>

>> >Let a = (aj)_j be a sequence of positive integers.
>> >
>> >Define [a1] = a1; [a1,a2] = a1 + 1/a2; [a1,a2,a3] = a1 + 1/(a2 + 1/a3);
>> > [a1,.. aj] = a1 + 1/(a2 + 1/(a3 +..+ 1/aj))..)
>> >
>> >Shome use the more visual notation
>> > a1 + 1/a2+ 1/a3+ ..+1/aj
>> >for [a1,.. aj].
>> >
>> >Let cj = [a1,.. aj].
>> >
>> >Assume j < k < n. How does one show
>> > |ck - c_n| < |cj - c_n|?

>>
>> It's enough to show this:
>>
>> (i) The numbers c_{j+1} - c_j are alternating
>> in sign and decreasing in absolute value.

>
>Done using 2x2 matrices. In fact, c_(j+1) = cj - (-1)^j / sj.s_(j+1)
>where rj, sj are positive coprime integers with cj = rj/sj
>and the r and s sequences are increasing.
>

>> (You have to figure out why what you want follows from (i)).
>
>That, in a nutshell, is my question.
>

>> Now (i) follows from (ii):
>>
>> (ii) The numbers c_{j+1} - c_j alternate in
>> sign. If j is odd then c_{j+2} > c_j,
>> while if j is even then c_{j+2} < c_j.

>
>Additionally, if j odd and k even, then cj < ck.
>

>> To prove (ii): Let d_j = [a_2,...,a_j], so that
>>
>> (*) c_j = 1 + 1/d_j.

>
>c2 = a1 + 1/d2. What did you intend?


c_j = a_1 + 1/d_j .

>> By induction you can assume (iii):
>>
>> (iii) The numbers d_{j+1} - d_j alternate in
>> sign. If j is odd then d_{j+2} < d_j,
>> while if j is even then d_{j+2} > d_j.
>>
>> And then it's not hard to show that (ii)
>> follows from (iii) and (*).
>>





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